Question

One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of the upper wire to that of the lower wire will be:\[\text{[Area of cross section of wire = 0.005 cm}^2, \, Y = 2 \times 10^{11} \, \text{N/m}^2, \, g = 10 \, \text{m/s}^2\text{]}\]

Answer 1

Correct Answer: 3

Given: - Mass of the upper load: \(2 \, \text{kg}\) - Mass of the lower load: \(1 \, \text{kg}\)

Step 1: Calculating the Tension in Each Wire

The tension in the upper wire (\(T_{\text{upper}}\)) is due to the combined weight of both loads:

\[ T_{\text{upper}} = (2 \, \text{kg} + 1 \, \text{kg}) \times g = 3 \times 10 = 30 \, \text{N} \]

The tension in the lower wire (\(T_{\text{lower}}\)) is due to the weight of the lower load:

\[ T_{\text{lower}} = 1 \times 10 = 10 \, \text{N} \]

Step 2: Calculating the Longitudinal Strain in Each Wire

The longitudinal strain (\(\epsilon\)) is given by:

\[ \epsilon = \frac{\text{Stress}}{Y} = \frac{T}{A \times Y} \]

Calculating the strain in the upper wire:

\[ \epsilon_{\text{upper}} = \frac{T_{\text{upper}}}{A \times Y} = \frac{30}{5 \times 10^{-7} \times 2 \times 10^{11}} \] \[ \epsilon_{\text{upper}} = \frac{30}{2 \times 10^{6}} = 1.5 \times 10^{-5} \]

Calculating the strain in the lower wire:

\[ \epsilon_{\text{lower}} = \frac{T_{\text{lower}}}{A \times Y} = \frac{10}{5 \times 10^{-7} \times 2 \times 10^{11}} \] \[ \epsilon_{\text{lower}} = \frac{10}{2 \times 10^{6}} = 0.5 \times 10^{-5} \]

Step 3: Calculating the Ratio of Longitudinal Strains

The ratio of the longitudinal strain of the upper wire to that of the lower wire is given by:

\[ \text{Ratio} = \frac{\epsilon_{\text{upper}}}{\epsilon_{\text{lower}}} = \frac{1.5 \times 10^{-5}}{0.5 \times 10^{-5}} = 3 \]

Conclusion: The ratio of longitudinal strain of the upper wire to that of the lower wire is \(3\).

Answer 2

Given: 

The system consists of masses connected with strings. The forces acting on the masses are \( g \) (gravitational acceleration), and the tensions are \( T_1 \) and \( T_2 \) for the corresponding strings.

Step 1: The net force on the system is given as zero, as the system is in equilibrium:

\[ \sum f_{\text{net}} = 0 \]

Step 2: The relation between the tensions \( T_1 \) and \( T_2 \) can be written as:

\[ T_2 = T_1 + 2g \]

Substitute the values for \( T_2 \) and \( T_1 \) in terms of the acceleration due to gravity (\( g \)):

\[ T_2 = g + 2g = 3g \]

Step 3: To calculate the strain and stress in the system:

The relationship between stress and strain is given by:

\[ (\text{stress})_L = y (\text{strain})_L \]

The strain is calculated by the formula:

\[ (\text{strain})_L = \frac{\text{stress}}{Y} = \frac{F}{YA} \]

Step 4: For the upper and lower strain components, we use the following relations:

\[ (\text{strain})_{\text{upper}} = \frac{T_2}{YA}, \quad (\text{strain})_{\text{lower}} = \frac{T_1}{YA} \]

Step 5: Substituting the known values for \( T_2 \) and \( T_1 \):

\[ (\text{strain})_{\text{upper}} = \frac{T_2}{YA} = \frac{3g}{YA}, \quad (\text{strain})_{\text{lower}} = \frac{T_1}{YA} = \frac{g}{YA} \]

Step 6: The strain ratio between the upper and lower components is then:

\[ \frac{(\text{strain})_{\text{upper}}}{(\text{strain})_{\text{lower}}} = \frac{3g}{g} = 3 \]